P2875 [USACO07FEB]牛的词汇The Cow Lexicon 三维dp 它慢,但它好写. 直接根据题意设三个状态: $f[i][j][k]$表示主串扫到第$i$个字母,匹配到第$j$个单词的第$k$位可以留下的最多字符数 当该位不选时,就传递上一位的数据$f[i][j][k]=f[i-1][j][k]$ 当该位可以匹配时: $if(a[i]==b[j][k]\&\&f[i-1][j][k-1])$$f[i][j][k]=max(f[i][j][k],f[i-1][j][k-1…

P2875 [USACO07FEB]牛的词汇The Cow Lexicon 题目描述 Few know that the cows have their own dictionary with W (1 ≤ W ≤ 600) words, each containing no more 25 of the characters 'a'..'z'. Their cowmunication system, based on mooing, is not very accurate; sometime…

题目描述 Few know that the cows have their own dictionary with W (1 ≤ W ≤ 600) words, each containing no more 25 of the characters 'a'..'z'. Their cowmunication system, based on mooing, is not very accurate; sometimes they hear words that do not make any…

https://daniu.luogu.org/problemnew/show/P2875 dp[i]表示前i-1个字符,最少删除多少个 枚举位置i, 如果打算从i开始匹配, 枚举单词j,计算从i开始往后,匹配完单词j的位置pos,删除的字母个数sum dp[pos]=min(dp[i]+sum) 如果不用i匹配,dp[i+1]=min(dp[i]) #include<cstdio> #include<cstring> #include<iostream> using…

传送门 f[i] 表示前 i 个字符去掉多少个 的最优解 直接暴力DP ——代码 #include <cstdio> #include <cstring> #include <iostream> ]; ], a[][]; inline int read() { , f = ; char ch = getchar(); ; ) + (x << ) + ch - '; return x * f; } inline int min(int x, int y) {…

题目 1633: [Usaco2007 Feb]The Cow Lexicon 牛的词典 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 401  Solved: 216[Submit][Status] Description 没有几个人知道,奶牛有她们自己的字典,里面的有W (1 ≤ W ≤ 600)个词,每个词的长度不超过25,且由小写字母组成.她们在交流时,由于各种原因,用词总是不那么准确.比如,贝茜听到有人对她说"browndcodw"…

P2863 [USACO06JAN]牛的舞会The Cow Prom 求点数$>1$的强连通分量数,裸的Tanjan模板. #include<iostream> #include<cstdio> #include<cstring> using namespace std; int min(int &a,int &b){return a<b?a:b;} #define N 10002 #define M 50002 int n,m,clo,df…

洛谷 2953 [USACO09OPEN]牛的数字游戏Cow Digit Game 题目描述 Bessie is playing a number game against Farmer John, and she wants you to help her achieve victory. Game i starts with an integer N_i (1 <= N_i <= 1,000,000). Bessie goes first, and then the two players…

本来分好组之后,就确定好了每个人要学什么,我去学数据结构啊. 因为前一段时间遇到一道题是用Lca写的,不会,就去学. 然后发现Lca分为在线算法和离线算法,在线算法有含RMQ的ST算法,前面的博客也写了.离线算法是基于DFS的Tarjan算法. 然后就打算去学一下Tarjan,因为以前也看过但是没看完,就打算学一下,因为Tarjan算法是图论的内容,然后就让图论选手教了我一下大环套小环的怎么推,然后就尴尬了. 我是学数据结构的,没有要去抢图论的内容学... 我也看了线段树了啊. 学完这个Tarj…

洛谷——P2863 [USACO06JAN]牛的舞会The Cow Prom 题目描述 The N (2 <= N <= 10,000) cows are so excited: it's prom night! They are dressed in their finest gowns, complete with corsages and new shoes. They know that tonight they will each try to perform the Round D…